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I don't know if this information will be helpful for those who are also trapped with M1 or for future debugging, but I found a rather consistent pattern of crashing and potentially a way to avoid it: Opusmodus on M1 always crash during threading. So be patient. Wait until the interpreter finishes its job and then execute the next block of code. This also applies to any manipulations on the GUI, like saving files, function search, etc. I know it sucks. But to make it suck less.. A side note. Earlier this year I helped upgrading a bunch of Python apps for M1, which also had issues of intermittent crash and it turned out the problem was also with threads. In that case the issue was solved when some problematic lines of code were removed (calling certain parent class methods during threading).

Congratulations, Stephane !! Best !

Small update to the final Conductor score and audio file. Best ! Stéphane

Yes, I just got it now. Thank you.

You should have received it by now. Thank you for your patience.

I also payed but I can't get or find the serial number. Please tell me how to get it.

Hi, here's a short score composed for string quartet. All the best to all OM users ! Stéphane stephaneboussuge · Méditation MeditationPourRené.opmo 01 - Conducteur - Méditation.pdf

Another good complet list can be found in function list under Tonality and Modes. SB.

Thank you very much ! It is exactly what I am looking for!

hth

Hi, where can I find the complete list of tonality-name presets, such as pentatonic-minor, messiaen modes, etc....?

Sorry, but I´m not the owner of this video... But is already in this link: https://youtu.be/RzPEORQNlZI Best !

Hello Julio, could you repost this video. It seems to be offline. Thanks, Achim

Cool !!! Thanks a lot !!! Best !

Done: (combination 3 '(0 1 2 3 4 5)) => ((0 1 2) (0 1 3) (0 1 4) (0 1 5) (0 2 3) (0 2 4) (0 2 5) (0 3 4) (0 3 5) (0 4 5) (1 2 3) (1 2 4) (1 2 5) (1 3 4) (1 3 5) (1 4 5) (2 3 4) (2 3 5) (2 4 5) (3 4 5)) (combination 3 '(0 1 2 3 4 5) :permute t) => ((0 1 2) (0 2 1) (1 0 2) (1 2 0) (2 0 1) (2 1 0) (0 1 3) (0 3 1) (1 0 3) (1 3 0) (3 0 1) (3 1 0) (0 1 4) (0 4 1) (1 0 4) (1 4 0) (4 0 1) (4 1 0) (0 1 5) (0 5 1) (1 0 5) (1 5 0) (5 0 1) (5 1 0) (0 2 3) (0 3 2) (2 0 3) (2 3 0) (3 0 2) (3 2 0) (0 2 4) (0 4 2) (2 0 4) (2 4 0) (4 0 2) (4 2 0) (0 2 5) (0 5 2) (2 0 5) (2 5 0) (5 0 2) (5 2 0) (0 3 4) (0 4 3) (3 0 4) (3 4 0) (4 0 3) (4 3 0) (0 3 5) (0 5 3) (3 0 5) (3 5 0) (5 0 3) (5 3 0) (0 4 5) (0 5 4) (4 0 5) (4 5 0) (5 0 4) (5 4 0) (1 2 3) (1 3 2) (2 1 3) (2 3 1) (3 1 2) (3 2 1) (1 2 4) (1 4 2) (2 1 4) (2 4 1) (4 1 2) (4 2 1) (1 2 5) (1 5 2) (2 1 5) (2 5 1) (5 1 2) (5 2 1) (1 3 4) (1 4 3) (3 1 4) (3 4 1) (4 1 3) (4 3 1) (1 3 5) (1 5 3) (3 1 5) (3 5 1) (5 1 3) (5 3 1) (1 4 5) (1 5 4) (4 1 5) (4 5 1) (5 1 4) (5 4 1) (2 3 4) (2 4 3) (3 2 4) (3 4 2) (4 2 3) (4 3 2) (2 3 5) (2 5 3) (3 2 5) (3 5 2) (5 2 3) (5 3 2) (2 4 5) (2 5 4) (4 2 5) (4 5 2) (5 2 4) (5 4 2) (3 4 5) (3 5 4) (4 3 5) (4 5 3) (5 3 4) (5 4 3))

Thanks, Janusz Here is the breakdown of the equivalent permutations for the example 012 021 102 120 201 210 013 031 103 130 301 310 014 041 104 140 401 410 015 051 105 150 501 510 023 032 230 203 302 320 024 042 204 240 402 420 025 052 205 250 502 520 034 043 304 340 403 430 035 053 305 350 503 530 045 405 450 504 540 054 125 152 215 251 512 521 123 132 213 231 312 321 124 142 214 241 412 421 135 153 315 351 513 531 134 143 314 341 413 431 145 154 415 451 514 541 234 243 324 342 423 432 235 253 325 352 523 532 245 254 425 452 524 542 345 354 435 453 534 543 The first column is the unique "seed" that gives the other permutations Just the seeds 012 013 014 015 023 024 025 034 035 045 125 123 124 135 134 145 234 235 245 345 Seeds above, arranged as complementary sets 012,345 013,245 014,235 015,234 023,145 024,135 025,134 034,125 035,124 045,123

I will change the combination to the definition and will add an additional function combination-permute which will return what you are looking for: ((0 1 2) (0 1 3) (0 1 4) (0 1 5) (0 2 1) (0 2 3) (0 2 4) (0 2 5) (0 3 1) (0 3 2) (0 3 4) (0 3 5) (0 4 1) (0 4 2) (0 4 3) (0 4 5) (0 5 1) (0 5 2) (0 5 3) (0 5 4 (1 0 2) (1 0 3) (1 0 4) (1 0 5) (1 2 0) (1 2 3) (1 2 4) (1 2 5) (1 3 0) (1 3 2) (1 3 4) (1 3 5) (1 4 0) (1 4 2) (1 4 3) (1 4 5) (1 5 0) (1 5 2) (1 5 3) (1 5 4) (2 0 1) (2 0 3) (2 0 4) (2 0 5) (2 1 0) (2 1 3) (2 1 4) (2 1 5) (2 3 0) (2 3 1) (2 3 4) (2 3 5) (2 4 0) (2 4 1) (2 4 3) (2 4 5) (2 5 0) (2 5 1) (2 5 3) (2 5 4) (3 0 1) (3 0 2) (3 0 4) (3 0 5) (3 1 0) (3 1 2) (3 1 4) (3 1 5) (3 2 0) (3 2 1) (3 2 4) (3 2 5) (3 4 0) (3 4 1) (3 4 2) (3 4 5) (3 5 0) (3 5 1) (3 5 2) (3 5 4) (4 0 1) (4 0 2) (4 0 3) (4 0 5) (4 1 0) (4 1 2) (4 1 3) (4 1 5) (4 2 0) (4 2 1) (4 2 3) (4 2 5) (4 3 0) (4 3 1) (4 3 2) (4 3 5) (4 5 0) (4 5 1) (4 5 2) (4 5 3) (5 0 1) (5 0 2) (5 0 3) (5 0 4) (5 1 0) (5 1 2) (5 1 3) (5 1 4) (5 2 0) (5 2 1) (5 2 3) (5 2 4) (5 3 0) (5 3 1) (5 3 2) (5 3 4) (5 4 0) (5 4 1) (5 4 2) (5 4 3)) The result of (combination 3 '(0 1 2 3 4 5)) will be: ((0 1 2) (0 1 3) (0 1 4) (0 1 5) (0 1 6) (0 2 3) (0 2 4) (0 2 5) (0 2 6) (0 3 4) (0 3 5) (0 3 6) (0 4 5) (0 4 6) (0 5 6) (1 2 3) (1 2 4) (1 2 5) (1 2 6) (1 3 4) (1 3 5) (1 3 6) (1 4 5) (1 4 6) (1 5 6) (2 3 4) (2 3 5) (2 3 6) (2 4 5) (2 4 6) (2 5 6) (3 4 5) (3 4 6) (3 5 6) (4 5 6))

According to the definition of the function: COMBINATION will return a list of all the combinations N elements in length from a list. There is something strange... For example: (combination 3 '(0 1 2)) gives ((0 1 2) (0 2 1) (1 0 2) (1 2 0) (2 0 1) (2 1 0)) and violate what the "unique combination" state you mentioned, since 0 1 2 is also (0 2 1) Looks like permutation... I´m looking for all n-permutations of a larger set, like 3 element permutations from a group of 6 elements (0 1 2 3 4 5). Maybe having the possibility of choosing the cardinality of permutations... Something like (permute 3 '(0 1 2 3 4 5)) should give the result I mentioned before (120 possbilities) Best, Julio

The result is correct. The function returns unique combination. 512 can't be as well 521.

Dear All, Recently, I used the function combination and got some strange result I did this (combination 3 '(0 1 2 3 4 5)) I got this result - 60 combinations (0 1 2) (0 1 3) (0 1 4) (0 1 5) (0 2 3) (0 2 4) (0 2 5) (0 3 4) (0 3 5) (0 4 5) (1 2 3) (1 2 4) (1 2 5) (1 3 4) (1 3 5) (1 4 5)(2 3 4) (2 3 5) (2 4 5) (3 4 5)(5 0 1) (5 0 2) (5 0 3) (5 0 4) (5 1 2) (5 1 3) (5 1 4) (5 2 3) (5 2 4) (5 3 4) ,(4 5 0) (4 5 1) (4 5 2) (4 5 3) (4 0 1) (4 0 2) (4 0 3) (4 1 2) (4 1 3) (4 2 3)(3 4 0) (3 4 1) (3 4 2) (3 5 0) (3 5 1) (3 5 2) (3 0 1) (3 0 2) (3 1 2) (2 3 0) (2 3 1) (2 4 0) (2 4 1) (2 5 0) (2 5 1) (2 0 1) (1 2 0) (1 3 0) (1 4 0) (1 5 0) There is something I did not get ? Best, Julio The expected result should be 6 x 5 x 4 = 120 Like this... 012 013 014 015 021 023 024 025 031 032 034 035 041 042 043 045 051 052 053 054 105 102 103 104 125 120 123 124 135 130 132 134 145 140 142 143 154 150 152 153 204 205 201 203 214 215 210 213 234 235 230 231 243 245 240 241 253 254 250 251 302 304 305 301 312 314 315 310 321 324 325 320 341 342 345 340 351 352 354 350 401 402 403 405 410 412 413 415 420 421 423 425 430 431 432 435 450 451 452 453 504 501 502 503 514 510 512 513 524 520 521 523 534 530 531 532 543 540 541 542

Thank you very much for this hint, good to know! best AO

Dorico Elements (Steinberg) works as well with microtonal xml files from Opusmodus, using Halion as a sound module.

Hi all, has anyone written a function that calculates the total duration of a score (including tempo changes and fermatas) in minutes and seconds? Achim

Perfect, thank you very much, all my Best, ciao Fabio